ICSE Class 10 Maths Probability Selina Solutions Guide
ICSE Class 10 Maths Probability: what this page covers
ICSE Class 10 Maths probability in Selina Concise Mathematics Chapter 25 is about finding the chance of an event by comparing favourable outcomes with all possible equally likely outcomes. This page gives a clean method sheet, step-by-step Selina-style worked solutions for the key probability questions, and original model questions for the common dice, coin, card and complement-event patterns used in this chapter.
The main rule is simple: if all outcomes are equally likely, then
P(E)=\frac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}.
Use this page to check the method, not only the final fraction. In probability, most errors happen before calculation: students count the wrong sample space, include repeated outcomes incorrectly, or forget to simplify the final answer.
Formula and method reference for ICSE Class 10 Maths probability
Before solving the exercise, fix the language. A random experiment is an action whose result is not known in advance. The sample space is the set of all possible outcomes. An event is the outcome or group of outcomes whose probability is required.
| Idea | What to write in the solution | Example |
|---|---|---|
| Basic probability | P(E)=\dfrac{n(E)}{n(S)} | For one die, S=\{1,2,3,4,5,6\}, so n(S)=6. |
| Complementary event | P(E')=1-P(E) | If P(E)=0.46, then P(E')=0.54. |
| Sure event | P(E)=1 | A number less than 8 on one die is certain. |
| Impossible event | P(E)=0 | Getting 8 on one ordinary die is impossible. |
| Range of probability | 0\leq P(E)\leq 1 | A probability cannot be negative or greater than 1. |
| Standard deck | 52 cards: 26 red, 26 black, 12 face cards | Red face cards =6, so probability =\dfrac{6}{52}. |
Concept snapshot: probability is a fraction of the sample space
Think of the sample space as a closed box containing all possible tickets. The numerator is the number of tickets that satisfy the condition. The denominator is the number of tickets in the box. If the question changes the box, the denominator changes. For example, choosing from the five vowels gives denominator 5, but choosing from the English alphabet gives denominator 26.
Exercise 25(A): worked Selina probability solutions
The following solutions use the standard school method: identify the sample space, count favourable outcomes, write the probability, and simplify the fraction. The question order follows the probability items in this Selina Chapter 25 page.
Question 1(a): Complementary events
Step 1: If A and B are complementary events, then exactly one of them occurs.
Step 2: The total probability of all possible outcomes is 1.
P(A)+P(B)=1.
Final answer: Option (c), P(A)+P(B)=1.
Question 1(b): Probability of selecting O from the vowels
Step 1: The English vowels are A,E,I,O,U. Hence, total possible outcomes =5.
Step 2: The favourable outcome is the single letter O. Hence, favourable outcomes =1.
P(\text{selecting }O)=\frac{1}{5}.
Final answer: Option (d), \dfrac{1}{5}.
Question 1(c): Even number greater than 4 on one die
Step 1: For one die, S=\{1,2,3,4,5,6\}. Therefore, n(S)=6.
Step 2: The even numbers greater than 4 in the sample space are \{6\}. Therefore, n(E)=1.
P(E)=\frac{n(E)}{n(S)}=\frac{1}{6}.
Final answer: Option (c), \dfrac{1}{6}.
Question 1(d): Letter of the word DELHI from the English alphabet
Step 1: One letter is drawn from the English alphabet, so n(S)=26.
Step 2: The letters in DELHI are D,E,L,H,I. They are 5 distinct letters.
P(\text{letter of }DELHI)=\frac{5}{26}.
Final answer: Option (b), \dfrac{5}{26}.
Question 1(e): Face card from a deck of 52 cards
Step 1: A standard deck has 52 cards.
Step 2: Face cards are jacks, queens and kings. There are 3 face cards in each suit and 4 suits.
n(E)=3\times 4=12.
P(\text{face card})=\frac{12}{52}=\frac{3}{13}.
Final answer: Option (a), \dfrac{3}{13}.
Question 2: A coin is tossed once
Step 1: For one coin toss, the sample space is S=\{H,T\}. Hence, n(S)=2.
Step 2: For getting a tail, E=\{T\}, so n(E)=1.
P(\text{tail})=\frac{1}{2}.
Step 3: Not getting a tail means getting a head, so the favourable set is \{H\}.
P(\text{not tail})=\frac{1}{2}.
Final answers: (i) \dfrac{1}{2}; (ii) \dfrac{1}{2}.
Question 3: Bag with white, black and red balls
Step 1: The bag contains 3 white, 5 black and 2 red balls.
n(S)=3+5+2=10.
Step 2: For a black ball, favourable outcomes =5.
P(\text{black})=\frac{5}{10}=\frac{1}{2}.
Step 3: For a red ball, favourable outcomes =2.
P(\text{red})=\frac{2}{10}=\frac{1}{5}.
Step 4: For a white ball, favourable outcomes =3.
P(\text{white})=\frac{3}{10}.
Step 5: Not red means white or black. Favourable outcomes =3+5=8.
P(\text{not red})=\frac{8}{10}=\frac{4}{5}.
Step 6: Not black means white or red. Favourable outcomes =3+2=5.
P(\text{not black})=\frac{5}{10}=\frac{1}{2}.
Final answers: (i) \dfrac{1}{2}; (ii) \dfrac{1}{5}; (iii) \dfrac{3}{10}; (iv) \dfrac{4}{5}; (v) \dfrac{1}{2}.
Question 4: One die thrown once
Step 1: For one die, S=\{1,2,3,4,5,6\}, so n(S)=6.
Step 2: Numbers greater than 4 are \{5,6\}.
P(\text{number greater than }4)=\frac{2}{6}=\frac{1}{3}.
Step 3: Numbers less than or equal to 4 are \{1,2,3,4\}.
P(\text{number }\leq 4)=\frac{4}{6}=\frac{2}{3}.
Step 4: Not greater than 4 means less than or equal to 4.
P(\text{not greater than }4)=\frac{4}{6}=\frac{2}{3}.
Final answers: (i) \dfrac{1}{3}; (ii) \dfrac{2}{3}; (iii) \dfrac{2}{3}.
Question 5: One card from a well-shuffled deck
Step 1: A standard deck contains 52 cards.
Step 2: Black cards are spades and clubs. There are 26 black cards.
P(\text{black card})=\frac{26}{52}=\frac{1}{2}.
Step 3: Not red means black, so favourable outcomes =26.
P(\text{not red})=\frac{26}{52}=\frac{1}{2}.
Step 4: Red cards are hearts and diamonds. There are 26 red cards.
P(\text{red card})=\frac{26}{52}=\frac{1}{2}.
Step 5: Face cards are J,Q,K in each of the 4 suits, so 12 face cards.
P(\text{face card})=\frac{12}{52}=\frac{3}{13}.
Step 6: Red face cards are J,Q,K of hearts and diamonds, so 6 cards.
P(\text{red face card})=\frac{6}{52}=\frac{3}{26}.
Final answers: (i) \dfrac{1}{2}; (ii) \dfrac{1}{2}; (iii) \dfrac{1}{2}; (iv) \dfrac{3}{13}; (v) \dfrac{3}{26}.
Question 6(i): Relation for complementary events
Step 1: Complementary events together cover the whole sample space and do not occur together.
Step 2: Therefore, their probabilities add up to 1.
P(A)+P(B)=1.
Final answer: P(A)+P(B)=1.
Question 6(ii): Probability of not happening of an event
Step 1: Given P(A)=0.46.
Step 2: Use the complement rule.
P(A')=1-P(A).
P(A')=1-0.46=0.54.
Final answer: The probability that event A does not happen is 0.54.
Question 7: Table tennis match between Geeta and Ritu
Step 1: Given P(\text{Ritu wins})=0.73.
Step 2: Assuming the match has only two possible winners, Geeta winning and Ritu winning are complementary events.
P(\text{Geeta wins})=1-P(\text{Ritu wins}).
P(\text{Geeta wins})=1-0.73=0.27.
Step 3: Not winning of Ritu is the same as Geeta winning under the same two-player assumption.
P(\text{Ritu does not win})=0.27.
Final answers: (i) 0.27; (ii) 0.27.
Question 8: Race between Mahesh and John
Step 1: Given P(\text{John loses})=0.54.
Step 2: Assuming the race is only between Mahesh and John and there is no tie, John losing means Mahesh winning.
P(\text{Mahesh wins})=0.54.
Step 3: John winning and John losing are complementary events.
P(\text{John wins})=1-P(\text{John loses}).
P(\text{John wins})=1-0.54=0.46.
Final answers: (i) 0.54; (ii) 0.46.
Question 9: Sure event, impossible event and range of probability
Step 1: A sure event always happens, so its probability is 1.
Step 2: An impossible event cannot happen, so its probability is 0.
Step 3: For any event E, probability lies from 0 to 1, both included.
0\leq P(E)\leq 1.
Final answers: (i) 1; (ii) 0; (iii) 0\leq P(E)\leq 1.
Question 10: Probability from one die
Step 1: The sample space for one die is S=\{1,2,3,4,5,6\}, so n(S)=6.
Step 2: For getting 5, favourable outcomes =\{5\}.
P(5)=\frac{1}{6}.
Step 3: A die has no face numbered 8, so favourable outcomes =0.
P(8)=\frac{0}{6}=0.
Step 4: Every face of a die is less than 8, so favourable outcomes =6.
P(\text{number less than }8)=\frac{6}{6}=1.
Step 5: Prime numbers on a die are \{2,3,5\}, so favourable outcomes =3.
P(\text{prime number})=\frac{3}{6}=\frac{1}{2}.
Final answers: (i) \dfrac{1}{6}; (ii) 0; (iii) 1; (iv) \dfrac{1}{2}.
Question 11: Even number, number between 3 and 8, and union event
Step 1: For one die, S=\{1,2,3,4,5,6\} and n(S)=6.
Step 2: Even numbers are \{2,4,6\}, so n(E)=3.
P(\text{even number})=\frac{3}{6}=\frac{1}{2}.
Step 3: The numbers between 3 and 8 in the die sample space are \{4,5,6\}, taking “between” in the usual exclusive sense.
P(\text{number between }3\text{ and }8)=\frac{3}{6}=\frac{1}{2}.
Step 4: Even numbers or multiples of 3 are \{2,3,4,6\}.
P(\text{even or multiple of }3)=\frac{4}{6}=\frac{2}{3}.
Final answers: (i) \dfrac{1}{2}; (ii) \dfrac{1}{2}; (iii) \dfrac{2}{3}.
Original worked examples for Selina Chapter 25 board practice
These examples are not copied from any source. They show how to handle the question types that usually follow the first basic exercise: two dice, cards, complementary events and choosing from a group.
Worked Example 1: Two dice and a sum condition
Question: Two dice are thrown together. Find the probability of getting a sum of 9.
Step 1: When two dice are thrown, each die has 6 outcomes.
n(S)=6\times 6=36.
Step 2: The ordered pairs giving sum 9 are (3,6),(4,5),(5,4),(6,3).
n(E)=4.
Step 3: Apply the probability formula.
P(\text{sum }9)=\frac{4}{36}=\frac{1}{9}.
Final answer: \dfrac{1}{9}.
Worked Example 2: Card probability with “not”
Question: One card is drawn from a well-shuffled pack of 52 cards. Find the probability that it is not a king.
Step 1: Total possible outcomes =52.
Step 2: There are 4 kings in a deck, so the number of cards that are not kings is 52-4=48.
P(\text{not king})=\frac{48}{52}=\frac{12}{13}.
Step 3: Check by complement rule.
P(\text{not king})=1-P(\text{king})=1-\frac{4}{52}=1-\frac{1}{13}=\frac{12}{13}.
Final answer: \dfrac{12}{13}.
Worked Example 3: Choosing a student from a group
Question: A class has 18 boys and 22 girls. One student is selected at random. Find the probability that the selected student is a girl.
Step 1: Find the total number of students.
n(S)=18+22=40.
Step 2: Favourable outcomes for selecting a girl =22.
P(\text{girl})=\frac{22}{40}=\frac{11}{20}.
Final answer: \dfrac{11}{20}.
Examiner’s mindset for ICSE Class 10 Maths probability answers
In probability questions, marks are usually earned by showing the count of total outcomes and favourable outcomes before writing the probability. A final fraction alone may not show that the sample space was counted correctly. For a deck question, write facts such as 52 total cards, 26 red cards, 12 face cards, or 6 red face cards before the probability line. For a die question, write S=\{1,2,3,4,5,6\} when the condition is likely to be misread.
For complement questions, the clearest method is P(E')=1-P(E). This avoids recounting and reduces decimal errors, especially in questions such as P(E)=0.73 or P(E)=0.46.
Common mistakes students make in probability
- Using the wrong denominator: If a letter is chosen from the vowels, the denominator is 5, not 26. If a letter is chosen from the English alphabet, the denominator is 26.
- Counting face cards as aces: In school probability, face cards are J,Q,K. Aces are not face cards. So face cards in a deck =12, not 16.
- Forgetting to simplify: Write \dfrac{12}{52}=\dfrac{3}{13}, not only \dfrac{12}{52}, unless the question asks for an unsimplified count.
- Misreading “or” in die questions: “Even number or multiple of 3” means the union of both sets. Count each outcome once: \{2,3,4,6\}, not 3+2=5 favourable outcomes.
- Ignoring assumptions in two-player questions: When the solution uses complements, it assumes only two possible results and no tie. State this assumption when needed.
Quick answer index
| Section | Question | Final answer |
|---|---|---|
| Exercise 25(A) | 1(a) | P(A)+P(B)=1 |
| Exercise 25(A) | 1(b) | \dfrac{1}{5} |
| Exercise 25(A) | 1(c) | \dfrac{1}{6} |
| Exercise 25(A) | 1(d) | \dfrac{5}{26} |
| Exercise 25(A) | 1(e) | \dfrac{3}{13} |
| Exercise 25(A) | 2(i), 2(ii) | \dfrac{1}{2}, \dfrac{1}{2} |
| Exercise 25(A) | 3(i) to 3(v) | \dfrac{1}{2}, \dfrac{1}{5}, \dfrac{3}{10}, \dfrac{4}{5}, \dfrac{1}{2} |
| Exercise 25(A) | 4(i) to 4(iii) | \dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3} |
| Exercise 25(A) | 5(i) to 5(v) | \dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2}, \dfrac{3}{13}, \dfrac{3}{26} |
| Exercise 25(A) | 6(i), 6(ii) | P(A)+P(B)=1,\ 0.54 |
| Exercise 25(A) | 7(i), 7(ii) | 0.27,\ 0.27 |
| Exercise 25(A) | 8(i), 8(ii) | 0.54,\ 0.46 |
| Exercise 25(A) | 9(i) to 9(iii) | 1,\ 0,\ 0\leq P(E)\leq 1 |
| Exercise 25(A) | 10(i) to 10(iv) | \dfrac{1}{6}, 0, 1, \dfrac{1}{2} |
| Exercise 25(A) | 11(i) to 11(iii) | \dfrac{1}{2}, \dfrac{1}{2}, \dfrac{2}{3} |
| Model practice | Two dice sum 9 | \dfrac{1}{9} |
| Model practice | Not a king | \dfrac{12}{13} |
| Model practice | Selecting a girl | \dfrac{11}{20} |
Related ICSE Class 10 Maths study links
Probability is usually revised with data handling and commercial mathematics topics because all three require careful reading of the question. For related practice, use ICSE Class 10 Banking solutions for formula substitution, Section Formula and Mid-Point Formula solutions for step-by-step coordinate work, and Heights and Distances solutions for trigonometry practice.
Authoritative references used
This page is aligned with the standard ICSE Class 10 Maths treatment of probability: sample space, favourable outcomes, complementary events, impossible events, sure events, die, coin and playing-card questions. For board-level reference, use the CISCE official website. For overlapping school-level probability definitions, the NCERT official website may also be used as a reference point. The worked solutions here are written independently for Selina Concise Mathematics Class 10, Chapter 25 Probability.
Frequently Asked Questions
What is the main formula used in ICSE Class 10 Maths probability?
The main formula is P(E)=\dfrac{n(E)}{n(S)}, where n(E) is the number of favourable outcomes and n(S) is the total number of equally likely outcomes.
How do I solve complement questions in Selina Chapter 25 Probability?
Use P(E')=1-P(E). For example, if the probability that an event happens is 0.46, then the probability that it does not happen is 1-0.46=0.54.
Are aces counted as face cards in Class 10 probability?
No. In standard ICSE Class 10 Maths card problems, face cards are jacks, queens and kings only. There are 12 face cards in a deck of 52 cards.
What is the probability of an impossible event and a sure event?
The probability of an impossible event is 0. The probability of a sure event is 1. For any event E, the probability always satisfies 0\leq P(E)\leq 1.
Why must probability answers be simplified?
Simplifying shows the final value clearly. For example, \dfrac{12}{52} for a face card should be reduced to \dfrac{3}{13}. The working may show the unsimplified count first, but the final answer should be simplified.